Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <stdio.h>
#include <vector>

int main()
{
    int n;
    scanf("%d", &n);
    if (n == 1) {
        return printf("0\n"), 0;
    }
    std::vector<int> p(n + 1), q(n + 1), t(n + 1), f(n + 1);
    for (int i = 1; i <= n; ++ i) {
        scanf("%d", &p[i]);
        q[p[i]] = i;
    }
    for (int i = n - 1; i >= 1; -- i) {
        if (t[i + 1] == 0) {
            if (q[i] > q[i + 1]) {
                t[i] = 1;
                f[i] = i + 1;
            } else {
                t[i] = 0;
            }
        } else {
            if (q[f[i + 1]] < q[i] && q[i] < q[i + 1] ||
                    q[i + 1] < q[f[i + 1]] && q[f[i + 1]] < q[i] ||
                    q[i] < q[i + 1] && q[i + 1] < q[f[i + 1]]) {
                t[i] = t[i + 1];
                f[i] = f[i + 1];
            } else {
                t[i] = t[i + 1] + 1;
                f[i] = i + 1;
            }
        }
    }
    printf("%d\n", t[1]);
}